有没有更好的方法来使用glob。Glob在python中获取多个文件类型的列表，如.txt， .mdown和.markdown?现在我有这样的东西:
projectFiles1 = glob.glob( os.path.join(projectDir, '*.txt') )
projectFiles2 = glob.glob( os.path.join(projectDir, '*.mdown') )
projectFiles3 = glob.glob( os.path.join(projectDir, '*.markdown') )
当前回答
你可以用这个:
project_files = []
file_extensions = ['txt','mdown','markdown']
for file_extension in file_extensions:
project_files.extend(glob.glob(projectDir + '*.' + file_extension))
其他回答
Glob返回一个列表:为什么不只是多次运行它并连接结果呢?
from glob import glob
project_files = glob('*.txt') + glob('*.mdown') + glob('*.markdown')
一句俏皮话，只是为了好玩。
folder = "C:\\multi_pattern_glob_one_liner"
files = [item for sublist in [glob.glob(folder + ext) for ext in ["/*.txt", "/*.bat"]] for item in sublist]
输出:
['C:\\multi_pattern_glob_one_liner\\dummy_txt.txt', 'C:\\multi_pattern_glob_one_liner\\dummy_bat.bat']
你可以用这个:
project_files = []
file_extensions = ['txt','mdown','markdown']
for file_extension in file_extensions:
project_files.extend(glob.glob(projectDir + '*.' + file_extension))
import os
import glob
projectFiles = [i for i in glob.glob(os.path.join(projectDir,"*")) if os.path.splitext(i)[-1].lower() in ['.txt','.markdown','.mdown']]
Os.path.splitext将返回filename & .extension
filename, .extension = os.path.splitext('filename.extension')
.lower()将字符串转换为小写
import glob
import pandas as pd
df1 = pd.DataFrame(columns=['A'])
for i in glob.glob('C:\dir\path\*.txt'):
df1 = df1.append({'A': i}, ignore_index=True)
for i in glob.glob('C:\dir\path\*.mdown'):
df1 = df1.append({'A': i}, ignore_index=True)
for i in glob.glob('C:\dir\path\*.markdown):
df1 = df1.append({'A': i}, ignore_index=True)
